Estimate the difference in energy between the first and second Bohr's orbit for a hydrogen atom. At what minimum atomic number , a transition from `n = 2` to `n = 1` energy level would result in the emission of X -rays with `lambda = 3.0 xx 10^(-8) m`? Which hydogen -like species does this atomic number correspond to ?

Explanation

The difference in energy is given by `Delta E`
`Delta E= 2.18 xx 10^(-18) (1)^(2)((1)/(1^(2)) - (1)/(2^(2))) J "atom"^(-1)`
`= 1.65 xx 10^(-18) J`
`-=1.65 xx 10^(-11) erg -= 10.2 eV`
For a H--like atom`lambda = 3.0 xx 10^(-8)m`
Hence the H-like atom is `He^(o+)`ion
To ionic `He^(o+)`ion,ionisation energy is `= - E_(1) IE= -(-13.6 xx Z^(2)) = + 54.4 eV`
`rArr overset(Delta E)underset((2 rarr 1)) = 2.18 xx 10^(-18) xx Z^(2) xx ((1)/(1^(2)) - (1)/(2^(2))) J`
`= E_("photon") = (hc)/(lambda)`
Solve to get `Z = 2`
The ionisation potential (IP) is the voltage difference required to generate this much energy)
`rArr IE = qV= e(IP) = 54.4 eV`
`rArr IP ("required") = 54.4 V`